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MISCELLANEOUS EXERCISE 1 (I) · Q224

Q.If x=a(cos⁡θ+θsin⁡θ)x=a(\cos\theta+\theta\sin\theta), y=a(sin⁡θ−θcos⁡θ)y=a(\sin\theta-\theta\cos\theta) then (d2ydx2)θ=π/4=\left(\dfrac{d^2y}{dx^2}\right)_{\theta=\pi/4}= (A) 82aπ\dfrac{8\sqrt2}{a\pi} (B) −82aπ-\dfrac{8\sqrt2}{a\pi} (C) aπ82\dfrac{a\pi}{8\sqrt2} (D) 42aπ\dfrac{4\sqrt2}{a\pi}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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dxdθ=a(−sin⁡θ+sin⁡θ+θcos⁡θ)=aθcos⁡θ,\frac{dx}{d\theta}=a(-\sin\theta+\sin\theta+\theta\cos\theta)=a\theta\cos\theta,

dydθ=a(cos⁡θ−cos⁡θ+θsin⁡θ)=aθsin⁡θ.\frac{dy}{d\theta}=a(\cos\theta-\cos\theta+\theta\sin\theta)=a\theta\sin\theta.

So dydx=aθsin⁡θaθcos⁡θ=tan⁡θ\dfrac{dy}{dx}=\dfrac{a\theta\sin\theta}{a\theta\cos\theta}=\tan\theta.

Differentiate again w.r.t. θ\theta: ddθ(tan⁡θ)=sec⁡2θ\dfrac{d}{d\theta}(\tan\theta)=\sec^2\theta, so

d2ydx2=sec⁡2θaθcos⁡θ=1aθcos⁡3θ.\frac{d^2y}{dx^2}=\frac{\sec^2\theta}{a\theta\cos\theta}=\frac{1}{a\theta\cos^3\theta}. …

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